Suppose $X$ is a set and $E_1,E_2,\dots,$ is a disjoint sequence of subsets of $X$ such that $\bigcup_{k=1}^{\infty}E_k=X$. Let $\mathcal{S}=\{\bigcup_{k\in\,K}E_k\,:\;K\subset\mathbb{Z}^{+}\}$. Prove that a function from $X$ to $\mathbb{R}$ is $\mathcal{S}$-measurable if and only if the function is constant on $E_k$ for every $k\in\mathbb{Z}^{+}$

I am stuck on the forward direction, don't see how a measurable function on disjoint set is a constant on each subset

Awesome! So, the question is: If$A-B = B-A$, then $A=B$. I proceeded by using proof by contradiction, like so: Assume $A \ne B$. Now, this means that there exists $x \in A$ but $x \notin B$. Furthermore, since we are given that $A-B=B-A$, we know that $x \in A-B$ and $x \in B-A$ (by definition of set equality), but this is a contradiction since $x \in A$ and $x \notin A$ at the same time, which goes against our initial assumption ($x \in A$)!

They are just minor things, like I think it is "more obvious" that $x\in B-A$ implies $x\in B$, but you get a similar contradiction either way. It's really the same thing, just looking at it upside down.

@Simple sorry I stopped responding: I think you need to go for a jog and clear your mind or something. Come back to this problem after a bit with a fresh canvas and try to work at it step by step.

Yeah, just come back to it later. Ping me if you need help and no one is giving it to you. The question is elementary enough that most people who know measure theory to some degree can probably help you walk through it.

First, we approach $\alpha_1$ and $\alpha_2$. We define these functions as equivalent to $f_i$, but restricted to $\text{ker}(g_i)$. We immediately note that if $a_i\in\text{ker}(g_i)$, then $h_i(g_i(a_i))=0$, and, by commutativity, we know that $h_i(g_i)=g_{i+1}(f_i)$ and so $f_i(a_i)\in\text{ker}(g_{i+1})$.

Thus, these $\alpha_i$ map from $\text{ker}(g_i)$ to $\text{ker}(g_{i+1})$. Further, since each $f_i$ is a well-defined homomorphism, we know that each $\alpha_i$ is a well-defined homomorphism

Since $f_2(f_1(a_1))=0$ for all $a_1\in A_1$, we know that $\text{Im}(\alpha_1)\subseteq\text{ker}(\alpha_2)$. Further, . . .

I supposedly have already finished the difficult part, in finding the bridging homomorphism and showing that it is well defined and a homomorphism

$f$ is measurable, by definition $f^{-1}(B)=\begin{cases}X\\\phi\end{cases}=\begin{cases}\cup_{k\in\,K}E_k\\\phi\end{cases}$ for every borel set of $R$, and $\cup_{k\in\,K}E_k$ is measurable which implies each $E_k$ is measurable

@Shootforthemoon I don't want to clutter up the comments on the post, so I am going to respond here. You said "Btw I mean it fits the definitions of course, but they come even after a simple proof such as x=0.999... - - >10x-x=9x=9 hence x=1."

This is not a proof. This is a heuristic, which is often used to justify the statement in an elementary setting. However, a lot of work needs to be done before this is actually a proof.

In particular, the notation $0.\overline{9}$ must be defined before you can start working with it.

in defense of that simple proof, I'll note that it's basically the same as the proof of the geometric series using stability and regularity of the summation method here

so it's an argument which works for any summation method with those as properties, and in that regard is neutral as to what method you use

(in that vein, I wonder if one could define $0.\overline{ab\ldots c}$ as the decimal number such that multiplying by 10 gives $a+0.\overline{b\ldots c}$ and adding 1 gives $1.\overline{ab\ldots c}$? but this gets back to the point of definition being important.)

sure. my point is that you don't actually need a lot in order to get 0.999... = 1, to the extent that a definition which didn't yield that would have to be in some sense unreasonable

(to say it differently, my intuition would be: one can choose to define 0.999... in such a way that it's 1, or you can define it in such a way that it's not a real number. but you can't define it in such a way that it's a real number other than 1.)

I think any objection I have to your statements is simply a matter of degree: I actually think that it takes a bit of work to define the real numbers in the first place. If we accept that the real numbers are there (or if we believe that they don't require much work to build), then one does not need "a lot" to get $0.\overline{9} = 1$.

@XanderHenderson sure. maybe the way I'd say it is this: I'd anticipate that it's not too hard to argue that, if 0.999... is to be a rational number, it's got to be 1. where things get dicey is replacing "rational number" with "real number"

@XanderHenderson Eventually I was asking what is the sense behind the definitions we gave and therefore the physical interpretation of the result we obtained. We got an equality between symbols. We should be able to see if that final result has some concrete meaning or is just abstract.

Who tells us that a limit is really accurate in defining for example real numbers? When the distance (theoretical or physical) between two numbers is zero, with the limit we conclude that the two objects are the same.